Toán Lớp 8: (2x-5) ² $-$ (x² +6x+9) = 0
Đề bài: tìm $x$
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: (2x-5) ² $-$ (x² +6x+9) = 0
Đề bài: tìm $x$
Comments ( 2 )
Giải đáp+Lời giải và giải thích chi tiết:
(2x-5)^2-(x^2+6x+9)=0
⇔(2x-5)^2-(x^2+2.x.3+3^2)=0
⇔(2x-5)^2-(x+3)^2=0
⇔(2x-5-x-3)(2x-5+x+3)=0
⇔(x-8)(3x-2)=0
⇔x-8=0⇒x=8
3x-2=0⇒3x=2⇒x=2/3
Vậy x={8,2/3}
(2x-5)^2-(x^2+6x+9)=0
->(2x-5)^2-(x+3)^2=0
->[(2x-5)-(x+3)].[(2x-5)+(x+3)]=0
->[2x-5-x-3].[2x-5+x+3]=0
->[(2x-x)-(5+3)].[(2x+x)+(3-5)]=0
->[x-8].[3x-2]=0
->\(\left[ \begin{array}{l}x-8=0\\3x-2=0\end{array} \right.\)
->\(\left[ \begin{array}{l}x=8\\3x=2\end{array} \right.\)
->\(\left[ \begin{array}{l}x=8\\x=\dfrac{2}{3}\end{array} \right.\)
Vậy x in {8; \frac{2}{3}}